Problem: You have found the following ages (in years) of all 5 tigers at your local zoo: $ 5,\enspace 10,\enspace 8,\enspace 4,\enspace 3$ What is the average age of the tigers at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 5 tigers at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{5 + 10 + 8 + 4 + 3}{{5}} = {6\text{ years old}} $ Find the squared deviations from the mean for each tiger. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $5$ years $-1$ years $1$ year $^2$ $10$ years $4$ years $16$ years $^2$ $8$ years $2$ years $4$ years $^2$ $4$ years $-2$ years $4$ years $^2$ $3$ years $-3$ years $9$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{1} + {16} + {4} + {4} + {9}} {{5}} $ $ {\sigma^2} = \dfrac{{34}}{{5}} = {6.8\text{ years}^2} $ The average tiger at the zoo is 6 years old. The population variance is 6.8 years $^2$.